Re: Picture with graph& question about area under a curve?

It's hard to help when you don't say what you did yourself! You say you are to use "36 equal subdivisions" and "use symmetry to make this problem simpler". Okay, I see 6 repetitions of the same thing, each of which is symmetric so I would divide one of them into 6 equal subdivisions and use the first 3.

Were you given a formula for this graph or only the graph itself? If only the graph itself, you will need to estimate the values at 0, 1/3, 2/3, and 1. That's hard to do with such a small graph. I might guess at 15, 4, 2, and 1 but those are only guesses. Using those heights for the rectangles, with width 1/3, we would have, using circumscribed rectangles, and the graph is decreasing on the left side of one loop, we use "right side" values (so the rectangles are completely outside the graph): 4(1/3)+ 2(1/3)+ 1(1/3)= (4+ 2+ 1)/3= 2 so that the entire area should be approximately 2(12)= 24.

That's roughly what you got. Why do you say it is too large? And what are those numbers at the end of your post?

If we were to use inscribed rectangles, we would use left side values: 15(1/3)+ 4(1/3)+ 2(1/3)= (15+ 4+ 2)/3= 7. Multiplying by 12 gives 84 for the entire area which is approximately what you got.

A better approximation (equivalent to the "trapezoid rule") would be the average of those two numbers: (84+ 24)/2= 108/2= 54.